1. **Problem statement:** We have a speed-time graph of Paul's cycling trip lasting 10 minutes. The graph has three segments: speed increases linearly from 0 to maximum speed $V$ in 4 minutes, remains constant at $V$ from 4 to 8 minutes, then decreases linearly back to 0 from 8 to 10 minutes. The total distance traveled is 1.68 km. We need to find the maximum speed $V$ in m/s. 2. **Formula and concept:** Distance traveled is the area under the speed-time graph. Since speed is in m/s and time is in minutes, convert time to seconds: $1 \text{ minute} = 60 \text{ seconds}$. 3. **Calculate total time in seconds:** $$10 \text{ minutes} = 10 \times 60 = 600 \text{ seconds}$$ 4. **Break the graph into three areas:** - Triangle from 0 to 4 minutes (0 to 240 seconds): area $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 240 \times V$ - Rectangle from 4 to 8 minutes (240 to 480 seconds): area $= \text{base} \times \text{height} = 240 \times V$ - Triangle from 8 to 10 minutes (480 to 600 seconds): area $= \frac{1}{2} \times 120 \times V$ 5. **Sum of areas = total distance:** $$\text{Distance} = \frac{1}{2} \times 240 \times V + 240 \times V + \frac{1}{2} \times 120 \times V = 120V + 240V + 60V = 420V$$ 6. **Convert total distance to meters:** $$1.68 \text{ km} = 1680 \text{ meters}$$ 7. **Set up equation and solve for $V$:** $$420V = 1680$$ $$V = \frac{1680}{420} = 4$$ 8. **Answer:** The maximum speed $V$ is $4$ m/s.